Final answer:
To find the value for which a matrix has one real eigenvalue of algebraic multiplicity two, calculate the characteristic polynomial and set its discriminant to zero to solve for the variable that results in a repeated root.
Step-by-step explanation:
The question is related to the eigenvalues of a matrix and their algebraic multiplicity. When a matrix has an eigenvalue with an algebraic multiplicity of two, it means that this eigenvalue is a repeated root of the matrix's characteristic polynomial. To determine the specific value that will give the matrix one real eigenvalue of algebraic multiplicity two, you usually need to calculate the characteristic polynomial and its roots.
An example of how to approach such a problem would be to first write down the matrix with a variable in place of the unknown entry. Next, you calculate the characteristic polynomial by subtracting λ times the identity matrix from the original matrix and then find the determinant of this new matrix. The resulting polynomial in λ is set to zero, and the solutions to this equation are the eigenvalues. If the matrix is supposed to have one real eigenvalue of algebraic multiplicity two, the characteristic polynomial would be a quadratic equation with a double root.